
\(x-2-\) \(\dfrac{x^2-10}{x+2}\)<...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(1.\) \(a.\) \(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\) \(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\) \(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\) \(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\) \(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\) \(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\) \(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\) \(=x-1\) \(b.\) \(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\) \(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\) \(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\) \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\) \(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\) \(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\) \(=\dfrac{2y}{\left(x-y\right)}\) Tương tự các câu còn lại 2) a) \(\dfrac{1}{x}.\dfrac{6x}{y}\) \(=\dfrac{6x}{xy}\) \(=\dfrac{6}{y}\) b) \(\dfrac{2x^2}{y}.3xy^2\) \(=\dfrac{2x^2.3xy^2}{y}\) \(=\dfrac{6x^3y^2}{y}\) \(=6x^3y\) c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\) \(=\dfrac{15x.2y^2}{7y^3.x^2}\) \(=\dfrac{30xy^2}{7x^2y^3}\) \(=\dfrac{30}{7xy}\) d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\) \(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\) \(=\dfrac{2y}{5x\left(x-y\right)}\) \(\dfrac{\left(x+y\right)2}{x2+xy}+\dfrac{\left(x-y\right)2}{x2-xy}=-\left(\dfrac{\left(x-y\right)2}{x2-xy}\right)+\dfrac{\left(x-y\right)2}{x2-xy}=0\) b: \(\dfrac{x^2-4x}{xy-4x-3y+12}+\dfrac{x-2}{y-4}\) \(=\dfrac{x\left(x-4\right)}{\left(y-4\right)\left(x-3\right)}+\dfrac{x-2}{y-4}\) \(=\dfrac{x^2-4x+x^2-5x+6}{\left(y-4\right)\left(x-3\right)}=\dfrac{2x^2-9x+6}{\left(y-4\right)\left(x-3\right)}\) c: \(=\dfrac{y^2}{\left(y-5\right)\left(x+1\right)}+\dfrac{2}{x+1}\) \(=\dfrac{y^2+2y-10}{\left(y-5\right)\left(x+1\right)}\) 1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\) \(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\) \(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\) \(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\) b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\) \(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\) \(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\) a) x(x - y) + y (x + y) = x2 – xy +yx + y2= x2+ y2 với x = -6, y = 8 biểu thức có giá trị là (-6)2 + 82 = 36 + 64 = 100 b) x(x2 - y) - x2 (x + y) + y (x2– x) = x3 – xy – x3 – x2y + yx2 - yx = -2xy Với x = \(\dfrac{1}{2}\), y = -100 biểu thức có giá trị là -2 . \(\dfrac{1}{2}\) . (-100) = 100. a)x(x-y)+y(x+y)=x2-xy+xy+y2=x2+y2 Tại x=-6 y=8 ta được : (-6)2+82=36+64=100 b) x(x2-y)-x2(x+y)+y(x2-x) =x3-xy-x3-x2y+x2y-xy=-2xy Tại x=\(\dfrac{1}{2}\) y=-100 ta được : (-2).\(\dfrac{1}{2}\).(-100)=-1.-100=100
